Page 143 - Physical Chemistry

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Chapter 4 With T held constant, a and V in (4.62) are functions of P only, and the integral is an
Material Equilibrium
ordinary integral. S for the process (P , T ) → (P , T ) equals S S .
1 1 2 2 a b
If the system undergoes a phase transition in a process, we must make separate
allowance for this change. For example, to calculate S for heating ice at 5°C and
1 atm to liquid water at 5°C and 1 atm, we use (4.61) to calculate the entropy change
for warming the ice to 0°C and for warming the water from 0°C to 5°C, but we
must also add in the entropy change [Eq. (3.25)] for the melting process. During melt-
ing, C dq /dT is infinite, and Eq. (4.61) doesn’t apply.
P P

EXAMPLE 4.4 S when both T and P change

Calculate S when 2.00 mol of water goes from 27°C and 1 atm to 37°C and
40 atm. Use data in Example 4.2 and neglect the pressure and temperature varia-
tions of C P,m , a, and V .
m
Equation (4.61) gives S 310 K (nC /T) dT, where the integration is at
a 300 K P,m
P P 1 atm. Neglecting the slight temperature dependence of C , we have
1 P,m
¢S 12.00 mol2375.3 J>1mol K24 ln 1310>3002 4.94 J>K
a
Equation (4.62) gives S 40 atm anV dP, where the integration is at T
b 1 atm m
T 310 K. Neglecting the pressure variation in a and V and assuming their
2 m
30°C values are close to their 37°C values, we have
1
3
¢S 10.000304 K 212.00 mol2118.1 cm >mol2139 atm2
b
3
3
3
0.43 cm atm>K 10.43 cm atm>K218.314 J2>182.06 cm atm2
0.04 J>K
¢S ¢S ¢S 4.94 J>K 0.04 J>K 4.90 J>K
b
a
Note the smallness of the pressure effect.
Exercise
Suppose that H O(l) goes from 29.0°C and 1 atm to 31.0°C and pressure P .
2 2
What value of P would make S 0 for this process? State any approximations
2
2
made. (Answer: 8.9 10 atm.)
Calculation of H and U
The use of Eqs. (4.30) and (4.48) in dH (
H/
T) dT (
H/
P) dP followed by
P T
integration gives
2 2
¢H C dT 1V TVa2 dP (4.63)
P
1 1
The line integrals in (4.63) are readily evaluated by using the path of Fig. 4.3. As usual,
separate allowance must be made for phase changes. H for a constant-pressure phase
change equals the heat of the transition.
U can be easily found from H using U H (PV). Alternatively, we
can write down an equation for U similar to (4.63) using either T and V or T and P as
variables.
Figure 4.4
Figures 4.4 and 4.5 plot u u and s s for H O(g) versus T and P, where u
tr,l tr,l 2 tr,l
Specific internal energy of H O(g) and s tr,l are the specific internal energy and specific entropy of liquid water at the triple
2
versus T and versus P. point (Sec. 1.5), and u U/m, s S/m, where m is the mass. The points on these